Nitrogen family Questions in English

(B) Phosphine $(PH_3)$ is a colorless,highly toxic gas that possesses a characteristic odor of rotten fish.

(D) Nessler's reagent is an alkaline solution of potassium tetraiodomercurate$(II)$,which has the chemical formula $K_2[HgI_4]$.
In this reagent,the active species responsible for the reaction with ammonia is the tetraiodomercurate$(II)$ ion,represented as $[HgI_4]^{2-}$.
When it reacts with ammonia,it forms a brown precipitate known as the iodide of Millon's base.

(A) The brown ring test is a common laboratory test for the detection of nitrate ions $(NO_3^-)$.
When a solution containing nitrate ions is treated with freshly prepared ferrous sulfate $(FeSO_4)$ solution and then concentrated sulfuric acid $(H_2SO_4)$ is added carefully along the sides of the test tube,a brown ring is formed at the interface.
The brown ring is due to the formation of the complex $[Fe(H_2O)_5(NO)]SO_4$.
In this reaction,the nitrate ion is reduced to nitric oxide $(NO)$,which then reacts with the ferrous ion to form the brown-colored complex.
Therefore,the gas responsible for this color formation is $NO$.

(C) When ammonia gas $(NH_3)$ reacts with hydrogen chloride gas $(HCl)$,they form solid ammonium chloride $(NH_4Cl)$ particles.
The reaction is: $NH_3(g) + HCl(g) \rightarrow NH_4Cl(s)$.
These solid particles appear as dense white fumes.

(C) When ammonia $(NH_3)$ reacts with excess chlorine $(Cl_2)$,the reaction produces nitrogen trichloride $(NCl_3)$ and hydrogen chloride $(HCl)$.
The balanced chemical equation is:
$NH_3 + 3Cl_2 \rightarrow NCl_3 + 3HCl$
Since chlorine is in excess,the reaction proceeds to form nitrogen trichloride,which is an explosive compound.

(B) The brown ring test is used to detect nitrate ions $(NO_3^-)$.
In this test,$Fe(II)$ ions react with nitric oxide $(NO)$ to form a brown-colored complex,$[Fe(H_2O)_5(NO)]^{2+}$.

(C) An amphoteric oxide is a substance that can react as both an acid and a base.
$Al_2O_3$ (Aluminum oxide) is a classic example of an amphoteric oxide because it reacts with both acids and bases to form salts and water.
$NO_2$ and $CO_2$ are acidic oxides.
Therefore,the correct option is $(C)$.

(B) The basic strength of nitrogen trihalides $(NX_3)$ depends on the availability of the lone pair of electrons on the nitrogen atom.
As the electronegativity of the halogen atom $(X)$ increases,it pulls the electron density away from the nitrogen atom due to the inductive effect.
The electronegativity order of halogens is $F > Cl > Br > I$.
Since fluorine is the most electronegative,it exerts the strongest electron-withdrawing effect on the nitrogen atom in $NF_3$.
This significantly reduces the availability of the lone pair on nitrogen,making $NF_3$ the weakest Lewis base among the nitrogen trihalides.

(D) The strength of oxyacids depends on the electronegativity of the central atom and the oxidation state.
$HNO_3$ is a strong acid because nitrogen is highly electronegative and the $N-O$ bond is highly polar,facilitating the release of $H^+$ ions.
In contrast,$H_3PO_4$,$H_3AsO_4$,and $H_3SbO_4$ are weak acids compared to $HNO_3$ because phosphorus,arsenic,and antimony are less electronegative than nitrogen.

(B) The acid anhydride is formed by the removal of water molecules from the acid.
For $H_3PO_4$: $2H_3PO_4 \rightarrow P_2O_5 + 3H_2O$.
For $HPO_3$: $2HPO_3 \rightarrow P_2O_5 + H_2O$.
Both $H_3PO_4$ (orthophosphoric acid) and $HPO_3$ (metaphosphoric acid) have $P_2O_5$ as their acid anhydride.

(D) Lewis base is a substance that can donate a lone pair of electrons.
In the group $15$ hydrides $(NH_3, PH_3, AsH_3, SbH_3)$,the central atom has one lone pair of electrons.
The availability of this lone pair for donation depends on the size of the central atom.
As we move down the group,the size of the central atom increases,and the lone pair occupies a larger orbital,making it less available for donation.
$NH_3$ has the smallest central atom $(N)$,so its lone pair is held most effectively and is most available for donation,making it the strongest Lewis base.

(A) The maximum oxidation state of an element is equal to the number of valence electrons in its outermost shell. For phosphorus $(Z = 15)$,the electronic configuration is $[Ne] 3s^2 3p^3$,so it has $5$ valence electrons,giving a maximum oxidation state of $+5$.
The minimum oxidation state is equal to the number of electrons required to achieve the nearest noble gas configuration. Phosphorus needs $3$ electrons to complete its octet,resulting in a minimum oxidation state of $-3$.
Therefore,the range of oxidation states for phosphorus is $-3$ to $+5$.

(B) The maximum oxidation state of nitrogen is determined by its valence electrons,which is $+5$.
The minimum oxidation state is determined by the number of electrons required to achieve the noble gas configuration,which is $-3$.
Therefore,nitrogen exhibits oxidation states ranging from $-3$ to $+5$.

(A) Nitrogen forms oxides in all oxidation states ranging from $+I$ to $+V$.

Oxide	Oxidation state of $N$
$N_2O$	$+1$
$NO$	$+2$
$N_2O_3$	$+3$
$NO_2$	$+4$
$N_2O_5$	$+5$

(C) The electronic configuration of the element with atomic number $Z = 33$ is $[Ar]^{18} \, 3d^{10} \, 4s^2 \, 4p^3$.
Since the last electron enters the $p$-orbital,it is a $p$-block element.
For $p$-block elements,the group number is calculated as $10 + (ns + np)$ electrons.
Group number $= 10 + 2 + 3 = 15$.

(C) The reaction of zinc with cold and very dilute nitric acid $(HNO_3)$ produces zinc nitrate and ammonium nitrate $(NH_4NO_3)$.
The balanced chemical equation is:
$4Zn + 10HNO_3 \rightarrow 4Zn(NO_3)_2 + NH_4NO_3 + 3H_2O$

(C) Silver nitrate is prepared by the reaction of silver with dilute nitric acid $(HNO_3)$ upon heating.
The chemical equation is:
$3Ag 4HNO_3 (\text{dilute}) \xrightarrow{\text{heat}} 3AgNO_3 NO 2H_2O$

(A) Phosphine $(PH_3)$ acts as a reducing agent and reduces silver nitrate $(AgNO_3)$ to metallic silver $(Ag)$.
The chemical reaction is:
$6AgNO_3 + PH_3 + 3H_2O \rightarrow 6Ag + 6HNO_3 + H_3PO_3$

(D) The Ostwald process involves the catalytic oxidation of ammonia $(NH_3)$ to nitric oxide $(NO)$.
The reaction is: $4NH_3(g) + 5O_2(g) \xrightarrow{Pt/Rh \text{ catalyst}} 4NO(g) + 6H_2O(g)$.
In this process,a platinum-rhodium $(Pt/Rh)$ gauze is used as a catalyst to facilitate the reaction.

(B) The Ostwald process involves the catalytic oxidation of ammonia $(NH_3)$ to nitric oxide $(NO)$.
The reaction is: $4NH_3(g) + 5O_2(g) \xrightarrow{Pt/Rh \text{ catalyst}} 4NO(g) + 6H_2O(g)$.
Here,a platinum-rhodium $(Pt/Rh)$ gauze is used as the catalyst to facilitate the reaction.

(B) When silver $(Ag)$ reacts with concentrated nitric acid $(HNO_3)$,it undergoes an oxidation-reduction reaction.
The reaction is as follows:
$Ag + 2HNO_3 (\text{conc.}) \rightarrow AgNO_3 + NO_2 + H_2O$
In this reaction,silver is oxidized to silver nitrate $(AgNO_3)$ and concentrated nitric acid is reduced to nitrogen dioxide $(NO_2)$ gas.

(C) The reaction between nitrogen and oxygen to form nitric oxide is an endothermic reaction that occurs at very high temperatures.
This reaction takes place in the internal combustion engine of automobiles at temperatures around $1483 \ K$.
Therefore,the correct option is $C$.

(D) The reaction between nitric oxide $(NO)$ and ozone $(O_3)$ is given by the equation:
$NO_{(g)} + O_{3(g)} \rightarrow NO_{2(g)} + O_{2(g)}$
Thus,the products formed are nitrogen dioxide $(NO_2)$ and oxygen $(O_2)$.

(A) In the atmosphere,nitrogen oxides exist primarily as $N_2O$ (nitrous oxide),$NO$ (nitric oxide),and $NO_2$ (nitrogen dioxide).
$N_2O$ is produced by bacterial action in soil.
$NO$ and $NO_2$ are produced by lightning and combustion processes in internal combustion engines.
Therefore,the correct set is $N_2O, NO, NO_2$.

(B) Nitrogen oxides like $NO$ and $NO_2$ are common atmospheric pollutants produced by combustion processes in automobiles and power plants.
$N_2O$ (nitrous oxide) is a naturally occurring gas in the atmosphere,primarily produced by microbial action in soil,and is not considered a common anthropogenic pollutant like $NO$ or $NO_2$.

(C) The chemical reaction between nitrogen dioxide $(NO_2)$ and ozone $(O_3)$ is given by the following equation:
$NO_{2(g)} + O_{3(g)} \rightarrow NO_{3(g)} + O_{2(g)}$
Thus,the products formed are nitrogen trioxide $(NO_3)$ and oxygen $(O_2)$.

(D) The metallic character of an element increases as we move down a group in the periodic table.
In Group $15$,the order of metallic character is $P < As < Sb < Bi$.
Therefore,$Bi$ is the most metallic element among the given options.

(D) Air is a mixture of gases,primarily $N_2$ $(78\%)$ and $O_2$ $(21\%)$,with small amounts of $CO_2$,water vapor,and noble gases.
To obtain pure nitrogen from air,oxygen is removed by passing air over heated copper or by other chemical methods,and carbon dioxide is removed by passing air through a potassium hydroxide $(KOH)$ solution.
Therefore,nitrogen is obtained by removing both oxygen and carbon dioxide.

(D) The thermal decomposition of orthophosphoric acid $(H_3PO_4)$ at $600\,^{\circ}C$ results in the formation of metaphosphoric acid $(HPO_3)$ and water vapor.
The chemical equation is: $H_3PO_4 \xrightarrow{600\,^{\circ}C} HPO_3 + H_2O$.
Therefore,the correct option is $(D)$.

(B) The azide ion $(N_3^-)$ has a linear structure with a bond angle of $180^{\circ}$.
It is considered a pseudohalogen because its chemistry resembles that of halide ions.
The formal oxidation state of nitrogen in $N_3^-$ is calculated as: $3x = -1$,so $x = -1/3$.
$N_3^-$ has $22$ electrons,while $NO_2$ has $23$ electrons,so they are not isoelectronic.

(B) The given reaction is the laboratory method for the preparation of phosphine $(PH_3)$ gas.
The balanced chemical equation is:
$P_4 + 3NaOH + 3H_2O \to PH_3 + 3NaH_2PO_2$
In this reaction,white phosphorus undergoes a disproportionation reaction where it is simultaneously oxidized and reduced to form phosphine $(PH_3)$ and sodium hypophosphite $(NaH_2PO_2)$.

(C) . Hydrazoic acid $(N_3H)$ is a fairly strong acid,which is stronger than $CH_3COOH$ but weaker than mineral acids.
$N_3H \to H^{+} + N_3^-$

(A) Ammonia $(NH_3)$ acts as a reducing agent when passed over heated copper$(II)$ oxide $(CuO)$.
The reaction is as follows:
$3CuO + 2NH_3 \xrightarrow{\text{heat}} N_2 + 3Cu + 3H_2O$
In this reaction,ammonia is oxidized to nitrogen gas $(N_2)$.

(A) . $N$ belongs to the second period and has only $2s$ and $2p$ orbitals available for bonding. It lacks $d$-orbitals,so it cannot expand its octet and thus cannot form $NCl_5$.
$P$ belongs to the third period and has vacant $3d$ orbitals available. This allows $P$ to undergo $sp^3d$ hybridization and expand its octet,enabling it to form both $PCl_3$ and $PCl_5$.

(B) When phosphine $(PH_3)$ gas is passed through an aqueous solution of copper sulphate $(CuSO_4)$,it reacts to form copper phosphide $(Cu_3P_2)$ as a precipitate.
The balanced chemical equation for this reaction is:
$3CuSO_4(aq) + 2PH_3(g) \to Cu_3P_2(s) + 3H_2SO_4(aq)$

(D) The inability of $ns^{2}$ electrons of the valence shell to participate in bonding when moving down the group in heavier $p$-block elements is known as the inert pair effect.
As a result,$Pb(II)$ is more stable than $Pb(IV)$,and $Sn(IV)$ is more stable than $Sn(II)$.
Since $Pb(IV)$ is unstable,it is easily reduced to $Pb(II)$,making $Pb(IV)$ a strong oxidizing agent.
Since $Sn(II)$ is unstable,it is easily oxidized to $Sn(IV)$,making $Sn(II)$ a strong reducing agent.

(A) The basicity of an oxoacid of phosphorus is determined by the number of $P-OH$ groups present in its structure.
Phosphinic acid $(H_3PO_2)$ contains one $P-OH$ group,making it a monoprotic acid.
Phosphonic acid $(H_3PO_3)$ contains two $P-OH$ groups,making it a diprotic acid.
Therefore,the correct statement is that phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid.

(D) The strong reducing behavior of $H_3PO_2$ (hypophosphorous acid) is attributed to its molecular structure.
In $H_3PO_2$,the phosphorus atom is bonded to two hydrogen atoms directly ($P-H$ bonds) and one hydroxyl group ($-OH$ bond).
Any oxyacid of phosphorus that contains at least one $P-H$ bond acts as a reducing agent.
Since $H_3PO_2$ contains two $P-H$ bonds,it exhibits strong reducing properties.

(B) Hypophosphorous acid,$H_{3}PO_{2}$,contains only one replaceable $H$-atom (that is attached to $O$,not with $P$ directly),so it is a monoprotic acid. Therefore,the statement that it is a diprotic acid is incorrect.

(B) As we move down group-$15$,the electronegativity of the central atom decreases.
Due to the decrease in electronegativity,the bond pairs of electrons move farther away from the central atom.
This results in a decrease in the $bp-bp$ (bond pair-bond pair) repulsion.
Consequently,the bond angle decreases from $NH_3$ to $SbH_3$.

(A) The stability of hydrides decreases from $NH_3$ to $BiH_3$ as we move down the group $15$. This is because the size of the central atom increases,leading to a weaker $M-H$ bond due to poor orbital overlap.
Therefore,the statement in option $A$ is incorrect.
Additionally,nitrogen cannot form $d\pi - p\pi$ bonds because it lacks $d$-orbitals in its valence shell.
The single $N-N$ bond is weaker than the $P-P$ bond due to high inter-electronic repulsion of the non-bonding electrons in the small $N$ atoms.
$N_2O_4$ exhibits resonance structures.

(A) Nitric oxide $(NO)$ is paramagnetic in the gaseous state because of the presence of one unpaired electron in its antibonding molecular orbital.
The molecular orbital configuration of $NO$ ($15$ electrons) is: $\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \sigma_{2p_z}^{2} \pi_{2p_x}^{2} \pi_{2p_y}^{2} \pi_{2p_x}^{*1}$.
Since it has an unpaired electron,it is paramagnetic,not diamagnetic.
Therefore,the statement that it is diamagnetic is incorrect.

(C) Nitrogen $(N_2)$ and oxygen $(O_2)$ are the primary components of the atmosphere.
They do not react under normal atmospheric conditions to form nitrogen oxides $(NO_x)$ because the reaction is highly endothermic and requires a very high temperature (e.g.,lightning or internal combustion engines) to overcome the high bond dissociation energy of the $N \equiv N$ triple bond.
Therefore,both the assertion and the reason are correct,and the reason provides the correct explanation for the assertion.

(C) $1.$ Orthophosphorous acid $(H_3PO_3)$: $3(+1) + x + 3(-2) = 0 \Rightarrow x = +3$
$2.$ Pyrophosphorous acid $(H_4P_2O_5)$: $4(+1) + 2x + 5(-2) = 0$ $\Rightarrow 2x = 6$ $\Rightarrow x = +3$
$3.$ Hypophosphoric acid $(H_4P_2O_6)$: $4(+1) + 2x + 6(-2) = 0$ $\Rightarrow 2x = 8$ $\Rightarrow x = +4$
$4.$ Pyrophosphoric acid $(H_4P_2O_7)$: $4(+1) + 2x + 7(-2) = 0$ $\Rightarrow 2x = 10$ $\Rightarrow x = +5$
Therefore,the pair with $+3$ oxidation state is Orthophosphorous and pyrophosphorous acids.

(B) The reaction of $Zn$ with dilute $HNO_3$ is given by:
$4 Zn + 10 HNO_3 (dil.) \longrightarrow 4 Zn(NO_3)_2 + 5 H_2O + N_2O$
Thus,dilute $HNO_3$ produces $N_2O$.
The reaction of $Zn$ with concentrated $HNO_3$ is given by:
$Zn + 4 HNO_3 (conc.) \longrightarrow Zn(NO_3)_2 + 2 H_2O + 2 NO_2$
Thus,concentrated $HNO_3$ produces $NO_2$.
Therefore,the products are $N_2O$ and $NO_2$ respectively.

(D) mixed anhydride is an oxide that reacts with water to produce two different acids,typically in different oxidation states.
$1$. $2ClO_2 + H_2O \rightarrow HClO_2 + HClO_3$ (Chlorous acid and Chloric acid).
$2$. $2NO_2 + H_2O \rightarrow HNO_2 + HNO_3$ (Nitrous acid and Nitric acid).
$3$. $2ClO_3$ is also considered a mixed anhydride as it disproportionates in water to form $HClO_3$ and $HClO_4$ (Chloric acid and Perchloric acid).
Therefore,all the given options are mixed anhydrides.

(C) Very pure nitrogen can be obtained by the thermal decomposition of sodium or barium azide.
$2 NaN_3 \xrightarrow{573 \ K} 2 Na + 3 N_2$
Azide salt of barium can be obtained in the purest form as the decomposition product contains solid $Ba$ as a by-product along with gaseous nitrogen,hence no additional step of separation is required.
$Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3 N_2$
Hence,option $C$ is correct.

(B) The reaction of $P_4$ with limited $O_2$ produces $P_4O_6$ (compound $X$).
The reaction of $P_4$ with excess $O_2$ produces $P_4O_{10}$ (compound $Y$).
In $P_4O_6$,each $P$ atom is bonded to three $O$ atoms via single bonds $(P-O)$.
In $P_4O_{10}$,each $P$ atom is bonded to three $O$ atoms via single bonds $(P-O)$ and one $O$ atom via a double bond $(P=O)$.
The presence of the $P=O$ double bond in $P_4O_{10}$ increases the oxidation state of $P$ and exerts a stronger inductive effect,which pulls electron density away from the $P-O$ single bonds.
This results in a shortening of the $P-O$ single bond length in $P_4O_{10}$ compared to $P_4O_6$.
Therefore,the bond length $d_{P-O}$ in $X$ $(P_4O_6)$ is greater than in $Y$ $(P_4O_{10})$,i.e.,$X > Y$.

(D) On heating,ammonium dichromate $(NH_4)_2Cr_2O_7$ undergoes thermal decomposition to produce nitrogen gas $(N_2)$,chromium$(III)$ oxide $(Cr_2O_3)$,and water $(H_2O)$.
The balanced chemical equation is: $(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} N_2 + Cr_2O_3 + 4H_2O$.